4-51SECTION 4.6..Integration by Substitution3932.When solving differential equations of the formdydt=f(y)for the unknown functiony(t), it is often convenient to makeuse of apotential functionV(y). This is a function such that−dVdy=f(y). For the functionf(y)=y−y3, find a potentialfunctionV(y). Find the locations of the local minima ofV(y)and use a graph ofV(y) to explain why this is called a “double-well” potential. Explain each step in the calculationdVdt=dVdydydt= −f(y)f(y)≤0.SincedVdt≤0, does the functionVincrease or decrease astime goes on? Use your graph ofVto predict the possible val-ues of limt→∞y(t). Thus, you can predict the limiting value ofthe solution of the differential equation without ever solvingthe equation itself. Use this technique to predict limt→∞y(t) ify=2−2y.3.Letfn(x)=2n+4n2x−12n≤x≤02n−4n2x0≤x≤12n0otherwiseforn=1,2,3,....For an arbitraryn, sketchy=fn(x)and show that1−1fn(x)dx=1. Compute limn→∞1−1fn(x)dx.Foranarbitraryx=0in[−2,2],computelimn→∞fn(x)andcompute1−1limn→∞fn(x)dx.Isitalwaystruethatlimn→∞1−1fn(x)dx=1−1limn→∞fn(x)dx?4.6INTEGRATION BY SUBSTITUTIONIn this section, we expand our ability to compute antiderivatives by developing a usefultechnique calledintegration by substitution.This method gives us a process for helpingto recognize a whole range of new antiderivatives.EXAMPLE 6.1Finding an Antiderivative by Trial and ErrorEvaluate2xex2dx.SolutionWe need to find a functionF(x) for whichF(x)=2xex2. You might betempted to guess that sincex2is an antiderivative of 2x,F(x)=x2ex2is an antiderivative of 2xex2. To see that this is incorrect, observe that, from the productrule,ddx(x2ex2)=2xex2+x2ex2(2x)=2xex2.So much for our guess. Before making another guess, look closely at the integrand.Notice that 2xis the derivative ofx2andx2already appears in the integrand, as theexponent ofex2. Further, by the chain rule, forF(x)=ex2,F(x)=ex2ddx(x2)=2xex2,which is the integrand. To finish this example, recall that we need to add an arbitraryconstant, to get2xex2dx=ex2+c.More generally, recognize that when one factor in an integrand is the derivative ofanother part of the integrand, you may be looking at a chain rule derivative.
394CHAPTER 4..Integration4-52Note that, in general, ifFis any antiderivative off, then from the chain rule, wehaveddx[F(u)]=F(u)dudx=f(u)dudx.From this, we have thatf(u)dudxdx=ddx[F(u)]dx=F(u)+c=f(u)du,(6.1)sinceFis an antiderivative off. If you read the expressions on the far left and the far rightsides of (6.1), this suggests thatdu=dudxdx.